Circuit Assitance

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Circuit Assitance

Postby alexanderfitu » January 10th, 2013, 3:00 am

Hi all

I am looking for some help with a motorbike brake light system. I have build some led arrays, 6 leds each in series, then in parallel for a total of 48 led's per board, both boards wired in series.

A picture says a thousand words so here is the "circuit" Diagram:

Image

Sorry for my crappy drawing, it was getting late and I was frustrated heh.

The reason for the resistors is that when the headlights are turned on the "tail" line is energised, at a lower current so that the LED arrays illumante dimmer, then when the brake is pressed, the "brake" line is energised, bypassing the resistors and getting brighter. This works when one array is connected, but is back feeding with more than one connected.

So when I connect the second LED strip (at the bottom of the diag), the LED arrays go out (when on tail mode) when on braking mode everything lights up as it should. Its my thinking that I need some diodes at A and B on the diagram, can anybody confirm if I am on the right track :)

Additionally when connecting the second led strip, when turning on the tail light, the brake lights up as well, I believe due to the lack of the diode, the 13v tail line is energising the brake line (albeit with reduced current because of the resistor)
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Re: Circuit Assitance

Postby k-ww » January 16th, 2013, 5:43 pm

Alexander:

First, please repost your diagram so that it shows the entire circuit.

Second, independant of your need for diodes, you are not taking
into account the minimum voltage that the LEDs would see -
for example, what if your battery has a shorted cell, and the motor
is stopped. You might have less than 10 Volts under these conditions,
and the LEDs might not light at all, and if they do, with reduced brightness.
Car Manufacturers use LED circuits that are designed so that they will
work even under low voltage conditions - 8 Volts or less, and use what
is called a constant current circuit so that the brightness of the LEDs
is independant of the battery voltage.
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Re: Circuit Assitance

Postby alexanderfitu » January 16th, 2013, 5:50 pm

Apologies, I did not realise it was cut off.

I have actually resolved the circuit by implementing this current circuit design:

http://i.imgur.com/CpIH5.png

In future, what would be better to use to supply a constanty current to the leds, but also prevent them from burning out at high voltage (as you say, so they work between 8v and 14v)?
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Re: Circuit Assitance

Postby k-ww » January 16th, 2013, 6:17 pm

Alaxander:

1) Pick a LED string that will work at 6V [say 3 diodes per string].

2) The constant current circuit will supply the same current to the
diode string regardless of what the supply voltage is, as long as it
is above the sum of the voltage across the diodes and what the
constant current circuit needs - a simple one is an NPN transistor
with a resistor in it's emitter lead to ground, and a string of 3 LEDs
connected to the transistor collector and +12V. If the transistor
has a turn on voltage of say .7V, and you have a 10 ohm resistor
in the emitter lead, when you supply say .8V between the
base of the transistor and ground, the transistor will drop .7V
and there will be .1V dropped arcoss the emitter resistor.
Ohms law tells us that .1V across 10 ohms gives 10mA,
so that 10mA flows thru the LEDs. It dosn't matter now if the 12V
changes to 8V or 20V, only 10mA flows thru the LEDs.

Change the voltage applied to the transistor's base to 1V, and 30mA
will flow thru the emitter resistor [and the LEDs].

Change the emitter resistor to 1 ohm, and either 100 or 300mA flows
thru the resistor [and LEDs]. connect a 470 ohm resistor to the battery
and connect two 1n4004 diodes in series to ground [cathodes towards ground],
and the juction of the 470 ohm resistor and the 'upper 1n4004 diode to
the base of the transistor and you have a 1.2 - 1.4V voltage reference
to drive the transistor. That gives you .5V to .7V across the emitter
resistor for 50/70mA or 500/700mA thru the LEDs. [note that you
need a power transistor that has a heatsink attached, because the
transistor has [50/70mA or 500/700mA times the voltage across
the transistor] in watts being disipated by the transistor.
[The higher the battery voltage, the more power is disipated by
the transistor also].

Don't forget to calculate the wattages the resistors are disipating.

Draw the circuit or google 'emitter follower' and think about it.

There are alternative circuits, but that is one way to build a constant
current circuit.
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